Math JEE Adv. 2025 paper 1

 To solve the problem, we first find the explicit forms of the functions 

$y_1(x)$, $y_2(x)$, and $y_3(x)$ by solving their respective differential equations. Then, we evaluate the given limit.

Step 1: Solve for $y_1(x)$

The differential equation is:

$$\frac{d{y}_1}{dx}-(\sin x{)}^2{y}_1=0,y_1(1)=5.$$

This is a separable equation. Rewrite it as:

$$\frac{d{y}_1}{y_1}=(\sin x{)}^{2}dx\mathrm{.}$$

Integrate both sides:

$$\ln y_1=\int (\sin x{)}^{2}dx+C\mathrm{.}$$

Using the identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$:

$$\ln y_1=\frac{1}{2}\int (1-\cos 2x)dx+C=\frac{x}{2}-\frac{\sin 2x}{4}+C\mathrm{.}$$

Exponentiate to solve for $y_1$:

$$y_1(x)=e^{\frac{x}{2}-\frac{\sin 2x}{4}+C}=A{e}^{\frac{x}{2}-\frac{\sin 2x}{4}},$$

where $A=e^C$. Use the initial condition $y_1(1)=5$:

$$5=A{e}^{\frac{1}{2}-\frac{\sin 2}{4}}\Rightarrow A=5e^{-\frac{1}{2}+\frac{\sin 2}{4}}\mathrm{.}$$

Thus:

$$y_1(x)=5e^{\frac{x-1}{2}-\frac{\sin 2x-\sin 2}{4}}\mathrm{.}$$

Step 2: Solve for $y_2(x)$

The differential equation is:

$$\frac{d{y}_2}{dx}-(\cos x{)}^2{y}_2=0,y_2(1)=\frac{1}{3}\mathrm{.}$$

Similarly:

$$\frac{d{y}_2}{y_2}=(\cos x{)}^{2}dx\mathrm{.}$$

Integrate both sides:

$$\ln y_2=\int (\cos x{)}^{2}dx+C=\frac{x}{2}+\frac{\sin 2x}{4}+C\mathrm{.}$$

Exponentiate:

$$y_2(x)=B{e}^{\frac{x}{2}+\frac{\sin 2x}{4}},$$

where $B=e^C$. Use the initial condition $y_2(1)=\frac{1}{3}$:

$$\frac{1}{3}=B{e}^{\frac{1}{2}+\frac{\sin 2}{4}}\Rightarrow B=\frac{1}{3}{e}^{-\frac{1}{2}-\frac{\sin 2}{4}}\mathrm{.}$$

Thus:

$$y_2(x)=\frac{1}{3}{e}^{\frac{x-1}{2}+\frac{\sin 2x-\sin 2}{4}}\mathrm{.}$$

Step 3: Solve for $y_3(x)$

The differential equation is:

$$\frac{d{y}_3}{dx}-\left(\frac{2-x^3}{x^3}\right)y_3=0,y_3(1)=\frac{3}{5e}\mathrm{.}$$

Rewrite it as:

$$\frac{d{y}_3}{y_3}=(\frac{2}{x^3}-1)dx\mathrm{.}$$

Integrate both sides:

$$\ln y_3=\int (\frac{2}{x^3}-1)dx=-\frac{1}{x^2}-x+C\mathrm{.}$$

Exponentiate:

$$y_3(x)=C^{\mathrm{\prime }}{e}^{-\frac{1}{x^2}-x},$$

where $C^{\mathrm{\prime }}=e^C$. Use the initial condition $y_3(1)=\frac{3}{5e}$:

$$\frac{3}{5e}=C^{\mathrm{\prime }}{e}^{-1-1}=C^{\mathrm{\prime }}{e}^{-2}\Rightarrow C^{\mathrm{\prime }}=\frac{3}{5}{e}^2\mathrm{.}$$

Thus:

$$y_3(x)=\frac{3}{5}{e}^{2-\frac{1}{x^2}-x}\mathrm{.}$$

Step 4: Compute the Product $y_1(x)y_2(x)y_3(x)$

Multiply the three solutions:

$$y_1(x)y_2(x)y_3(x)=5\cdot \frac{1}{3}\cdot \frac{3}{5}{e}^{(\frac{x-1}{2}-\frac{\sin 2x-\sin 2}{4})+(\frac{x-1}{2}+\frac{\sin 2x-\sin 2}{4})+(2-\frac{1}{x^2}-x)}\mathrm{.}$$

Simplify the exponents:

$$=e^{(x-1)+2-\frac{1}{x^2}-x}=e^{1-\frac{1}{x^2}}\mathrm{.}$$

Step 5: Evaluate the Limit

The limit to compute is:

$$\underset{x\to 0^{+}}{\lim }\frac{y_1(x)y_2(x)y_3(x)+2x}{e^{3x}\sin x}=\underset{x\to 0^{+}}{\lim }\frac{e^{1-\frac{1}{x^2}}+2x}{e^{3x}\sin x}\mathrm{.}$$

As $x\to 0^{+}$, $e^{1-\frac{1}{x^2}}\to 0$ (since $-\frac{1}{x^2}\to -\mathrm{\infty }$), and $e^{3x}\sin x\approx x$ (since $\sin x\approx x$ and $e^{3x}\approx 1$). Thus:

$$\frac{0+2x}{x}=2.$$

Final Answer:

$$\boxed{{\displaystyle 2}}$$